What are the answers to -4+sin(x)=-8+sqrt2/2

Question

I need to know how to find the two answers between 0 and 2 pi. I need the answers in degrees, not radians. there should be two answers

Answer 1

 sin x = (e^ix – e^-ix)/2i

Let sin x = -4 + sqrt2/2 = k = (e^ix – e^-ix)/2i

2ik = (e^ix – e^-ix)

(e^ix – e^-ix) –2ik = 0

e^2ix – 2ike^ix –1 = 0

Let u = e^ix

u^2 – 2iku – 1 = 0, u = (2ik ± (((-2ik)^2 – 4(1)(-1))^(1/2))/2

(-2.586i ± 1.64)/2

u=  e^ix   ≈  -.4735i, -2.113i

e^ix = -.4735i

ix = ln(-.4735i)

ix = ln(.4735) - (pi)/2

x = -i ln(.4735) –  (pi)/2

x = -i(-.748) +  (pi)/2

x = -1.57 + .748i

|x| = 1.739, Θ = arctan (.748/-1.57) = 154.5° OR 334.5º

AND

e^ix = -2.113i

ix = ln (-2.113i)

ix = ln (2.113) – ipi/2

x = -i ln (2.113) + (pi)/2

x = -1.57 - .748i

|x| = 1.739, Θ = 25.5° OR  205.5° 

 

Answer 2

l sin(x) l ≤ 1 ---> -1 ≤ sin(x) ≤ 1

sin (x) = -4 + (√2)/2 ---> sin(x) ≈ - 3.3 < - 1 , sin(x) is not exist there

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[0, 360^o]
sin x = (√2)/2 ----> x₁ = 45^o and x₂ = 135^o